A 3-phase induction motor runs at a speed of 1485 rpm at no-load and at 1350 rpm at fullload when supplied from a 50-Hz, 3-phase line. (a) How many poles does the motor have? (b) What is the % slip at no-load and at fullload? (c) What is the frequency of rotor voltages at no-load and at full-load? (d) What is the speed at both no-load and fullload of: (i) the rotor field with respect to rotor conductors, (ii) the rotor field with respect to the stator, and (iii) the rotor field with respect to the stator field.

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Solution: given dataNo Load speed N_{n L}=1485 \mathrm{rpm}F्याl load speed N_{F}=N_{r}=1350 r5msuptiy srequancy z=50 \mathrm{~Hz}.\rightarrow Induction paotor m 0-laad speed is approximaterz exyal to syrchrorous spered of the Motor.Hence Assurne syrchrorout speed N_{s}=1500a) symchronous speed, Ns =\frac{120 \mathrm{f}}{P}\begin{aligned}1500 & =\frac{120+50}{P} \\P & =4\end{aligned}Number of poles of the retor p=4\begin{array}{l}\text { SITp at } 200-2 \text { od, } S_{n L}=\frac{N s-N_{n L}}{N s} \\S_{n L}=\frac{1500-1485}{1500}=0.01 \\\text { Slip(SnL) }=0.01 \text { or } 1 \%\end{array}\begin{array}{l}\text { ii) Slip at fNN-Load } S_{F L}=\frac{N_{C}-N^{\prime}}{N_{S}} \\=\frac{1500-1350}{1500}=0-1 \\\therefore \text { Fuा } \operatorname{loa} \text { siाp }\left(5 \mathrm{SL}^{2}\right)=0.108101 . \\\end{array}c) fresurnczy of retor voltege = sfi) frequencry of reter voltage att koaload is\begin{array}{l}f_{n L}^{\prime}=s_{n L} f=0.01 \times 50 \\f_{n L}^{\prime}=0.5+z\end{array}ii) frequenciy कt rosor voltage at foll had isf_{f}=5 f_{L} \times f=0-1 \times 5=5+zd) speed gis the rotor field with respect to He rotor condvctorls meant speed of the rotatimg magretic field with respect to \gamma 0 tor.i) \therefore steea of the rotos tield with rispert to the rotor corducors at 10+ oad =\frac{205 n L f}{P}=\frac{120 \times 0.01 \times 50}{2+}=15 \times 9 m\begin{array}{l}\text { ii) Slip at fNN-Load } S_{F L}=\frac{N_{C}-N^{\prime}}{N_{S}} \\=\frac{1500-1350}{1500}=0-1 \\\therefore \text { Fuा } \operatorname{loa} \text { siाp }\left(5 \mathrm{SL}^{2}\right)=0.108101 . \\\end{array}c) fresurnczy of retor voltege = sfi) frequencry of reter voltage att koaload is\begin{array}{l}f_{n L}^{\prime}=s_{n L} f=0.01 \times 50 \\f_{n L}^{\prime}=0.5+z\end{array}ii) frequenciy कt rosor voltage at foll had isf_{f}=5 f_{L} \times f=0-1 \times 5=5+zd) speed gis the rotor field with respect to He rotor condvctorls meant speed of the rotatimg magretic field with respect to \gamma 0 tor.i) \therefore steea of the rotos tield with rispert to the rotor corducors at 10+ oad =\frac{205 n L f}{P}=\frac{120 \times 0.01 \times 50}{2+}=15 \times 9 m ...